[tex]\sf\fbox\red{Answer:-}[/tex]
(2+3)²
=(5)²
=25
[tex]\small\fbox{\blue{\underline{mαrk \; mє \; вrαínlíєѕt \; plєαѕє ♥}}}[/tex]
What is the length of diagonal LN?
12.6 in
25.2 in
22.8 in
20.4 in
Answer:
50.7016
Step-by-step explanation:
1.Area of rhombus is l^2
area of rhombus=1/2*d1*d2
Answer:
the answer is C.25.2
Step-by-step explanation:
i took the quiz and got it right.
A server makes $200 a week working at a local restaurant. The server saves 35% of the $200 weekly earnings each week for 4 weeks.
If the server saves 7/8 of the 5th week’s $200 earnings, how much money will the server have saved in total after the 5 weeks?
.$245
.$305
.$315
.$455
The total amount of money the server would have saved after 5 weeks is $455.
What is the total amount saved?The first step is to determine the total amount of money saved for the first four weeks.
Total amount of money saved in the first four weeks = 35% x (200 x 4) = $280
Amount saved on the 5th week = 7/8 x $200 = $175
Total savings = $175 + $280 = $455
To learn more about fractions, please check: https://brainly.com/question/1301963
can someone help me please
Answer:
C. d = C/pi
Step-by-step explanation:
lol nothing really to explain but it's saying diameter equals circumference divided by pi
Hope this helps dude !
on saturday mark have 20 apples and his mom took 2
What is the volume of the sphere?
Use \mathrm{\pi } = 3 . 14π=3.14 and round your answer to the nearest hundredth.
Answer:
cubic inches
The volume of the sphere that has a radius of 3 inches is: 113.1 in.³
What is the Volume of a Sphere?Volume of a sphere is found using the formula, V = 4/3 πr³, where r is the radius of the sphere.
Given the following:
π = 3.14r = 3 inchesPlug in the values into V = 4/3 πr³:
Volume of the sphere = 4/3 (3.14)(3)³
Volume of the sphere = 113.1 in.³
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if a:b=3:4 find the value of 15a+9b:16a+5b
Answer :
81 : 68⠀
Step-by-step explanation:
⠀
[tex] {\longrightarrow \: \sf \: Let \: \: a : b\: = 3: 4 = \dfrac{a}{b} \: = \: \dfrac{3}{4} = k}[/tex]
⠀
Therefore,
⠀
[tex] \dashrightarrow \qquad \sf \: a = 3k[/tex]
[tex] \dashrightarrow \qquad \sf \: b = 4k[/tex]
⠀
Now, Substituting the values :
⠀
[tex]\longrightarrow \qquad \sf \: 15a+9b:16a+5b[/tex]
⠀
[tex]\longrightarrow \qquad \sf \: \dfrac{15(3k)+9(4k)}{16(3k)+5(4k)}[/tex]
⠀
[tex]\longrightarrow \qquad \sf \: \dfrac{45k+36k}{48k+20k}[/tex]
⠀
[tex]\longrightarrow \qquad \sf \: \dfrac{81k}{68k}[/tex]
⠀
[tex]\longrightarrow \qquad \sf \: \dfrac{81 \cancel{k}}{68 \cancel{k}}[/tex]
⠀
[tex]\longrightarrow \qquad \sf \: \dfrac{81}{68}[/tex]
⠀
[tex]\longrightarrow \qquad \bf \: {81} : {68}[/tex]
⠀
Therefore ,
⠀
[tex]{ \pmb{ \sf{\longrightarrow \qquad \: 15a+9b:16a+5b = 81 : 68}}}[/tex]
⠀
Ashley has ridden 32 miles of a bike course. The course is 80 miles long. What percentage of the course has Ashley ridden so far?
Answer:
40%
Step-by-step explanation:
divide 80 by 100 80/100= 0.8 then divide 32 by 0.8 to get 40%
They are ordering 120 pounds of seafood for the week. How many should be fish?
Answer:
43.2 pounds of fish :)
Step-by-step explanation:
120 x 0.36 = 43.2
Have a great day!!
Please rate and mark brainliest!!!
A candy distributor needs to mix a 40% fat-content chocolate with a 50% fat-content chocolate to create 100 kilograms of a 41% fat-content chocolate. How many kilograms of each kind of chocolate must they use? A) Write an equation that uses the information as it is given above that can be used to solve this problem. Use x as your variable to represent the quantity of 40% fat-content chocolate. Equation: B) Answer: They must mix ?Kilograms of the 40% chocolate and ?Kilograms of the 50% chocolate.
Answer:
The kilograms of chocolate for the 40% fat content (1st chocolate) = 90kg
The kilograms of chocolate for the 50% fat content (2nd chocolate) = 10kg
Step-by-step explanation:
A candy distributor needs to mix a 40% fat-content chocolate with a 50% fat-content chocolate to create 100 kilograms of a 41% fat-content chocolate. How many kilograms of each kind of chocolate must they use?
Let us represent
Kilograms of first chocolate = x
Kilograms of second chocolate = y
A candy distributor needs to mix a 40% fat-content chocolate with a 50% fat-content chocolate to create 100 kilograms of a 41% fat-content chocolate.
Hence:
40% × x + 50% × y = 41% × 100
0.40x + 0.50y = 0.41× 100
0.40x + 0.50y = 41.......Equation 1
x + y = 100kg ......Equation 2
x = 100 - y
Substitute 100 - y for x
0.40(100 - y) + 0.50y = 41
40 - 0.40y + 0.50y = 41
Collect like terms
- 0.40y + 0.50y = 41 - 40
0.10y = 1
y = 1/0.1
y = 10.
Therefore, the kilograms of chocolate for the 50% fat content (2nd chocolate) = 10kg
Remember:
x = 100 - y
x = 100kg - 10kg
x = 90kg
The kilograms of chocolate for the 40% fat content (1st chocolate) = 90kg
If 1cm represents 5m, what would be an area of 6 square cm. represent?
Answer:
150 m²
Step-by-step explanation:
Given 1 cm represents 5 m
⟹1 cm=5 m
given area =6 cm² =6×(5 m)² =6×25 m²
=150 m²
area of 6 cm² represents 150 m²
What is the answer to this question I can’t work it out Ni=n02(yi-yo)/t2
Changing the subject of an equation involves solving for another variable in the equation
The equation of t is [tex]t=\frac{\sqrt{y_i-y_o}}{N}[/tex]
How to change the subject of the equationThe equation is given as:
[tex]N=\sqrt[2]{\frac{y_i-y_o}{t^2}}[/tex]
Rewrite the equation as:
[tex]N=\sqrt{\frac{y_i-y_o}{t^2}}[/tex]
Evaluate the square root of t^2
[tex]N=\frac{\sqrt{y_i-y_o}}{t}[/tex]
Multiply both sides by t
[tex]Nt=\sqrt{y_i-y_o}[/tex]
Make t the subject, in the above equation
[tex]t=\frac{\sqrt{y_i-y_o}}{N}[/tex]
Hence, the equation of t is [tex]t=\frac{\sqrt{y_i-y_o}}{N}[/tex]
Read more about subject of formula at:
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A shipping container is in the form of a right rectangular prism, with dimensions of
20 ft by 8 ft by 7 ft 9 in. If the container holds 409 cubic feet of shipped goods, what
percent is full? Round your answer to the nearest whole number if necessary.
Answer:
465
Step-by-step explanation:
I just did it
A total of 360 tickets were sold for a school play. They were either adult tickets or student tickets. The number of student tickets sold was two times the number of adult tickets sold. How many adult tickets were sold? (Please help)
Answer:
120 ----------adult tickets sold.240 (children tickets sold)Step-by-step explanation:
a total of 360 tickets were sold for the school play. they were either adults tickets or student tickets. the number of student tickets sold was two times the number of adult tickets sold. how many adult tickets were sold?
Answer by chibabykalu1 (Show Source): You can put this solution on YOUR website!
x, =student tickets
y, =adult tickets
-----
=> x + y = 360 --------(1) x = 2y. ----------(2)exact translation of the description into a system of equations. Solve for the value of y.
The process can be, solve for x in terms of y, already seen as THE second equation of the system; substitute into the other equation, and solve the result for y, and compute.
=> {2y} + y = 360=> 2y + y = 360=> 3y = 360=> (3y)/3 = +360)/4=> y = 120 ----------adult tickets sold.Substitute y = 120 into equation 2
=> x = 2y=> x = 2(120)=> x = 240 (children tickets sold)Complete two expressions that have the same product as 3 x 5 . 12
__________ x
1
12 __________ x
3
12 hurry up please
At a birthday party for 20 people each person could order a hot dog or chips. If there were four times as many hot dogs as chips ordered, calculate how many hot dogs and how many chips were bought.
Call the amount of chips ordered : x
-> The amount of hot dogs ordered : 4x
Since there are a total of 20 chips and hot dogs orders, x + 4x = 20 -> 5x = 20
-> x = 20/5 = 4
-> 4x = 4 x 4 = 16
So, there are 16 hot dogs and 4 chips ordered.
Answer:
16 hot dogs
4 chips
Step-by-step explanation:
Let h = number of hot dogs ordered
Let c = number of chips ordered
Given:
Total number of orders = 20Four times as many hot dogs as chips ordered⇒ h + c = 20
⇒ h = 4c
Substitute h = 4c into h + c = 20 and solve for c:
⇒ 4c + c = 20
⇒ 5c = 20
⇒ c = 4
Substitute found value for c into h = 4c and solve for h:
⇒ h = 4 × 4
⇒ h = 16
Therefore, 16 hot dogs and 4 chips were ordered.
Help?? I'm so confused
Answer:
0.10
Step-by-step explanation:
If one meter is 100 centimeters and you're trying to find how much of a meter is in 10 centimeters all you need to do is divide the meter by 10, just like they did to get the amount of centimeters.
Mary is saving money to buy a game.So far she has saved $36,which is three-fourths of the total cost of the game.How much does the game cost?
Answer:
48 i think
Step-by-step explanation:
Answer:
$48
Step-by-step explanation:
What value of x makes the equation true?
-4( x - 3) = -28
Answer: x = 10
Step-by-step explanation: hope this helps
show that
[tex]3 \frac{3}{4} \div 1 \frac{1}{3 } = 5 \frac{1}{12} [/tex]
write down all the steps in your working
3x+7+56=90 what is this
Answer:
X=9
Hope this helps! :)
Write the sum of O.31 (repeating 31 )and 0.125 (repeating 125)
Answer: 0.435 repeating
Step-by-step explanation:
PLEASE HELP DUE SOON! - 40 POINTS + BRAINLIEST
(2 attachments)
Answer:
Step-by-step explanation:
B
A
A
3.
Calculate 7.561 – 2.37
Answer:
7.561 – 2.37=5.191 if you wanted me to show work, comment and ill change it.
Step-by-step explanation:
please help thank youuu
Answer:
What color does the letter purple taste like?
Step-by-step explanation:
Find the Derivative y’ implicitly.
[tex]e^{x^2y} - e^y = e^y(e^{x^2} - 1) = x[/tex]
We should ISOLATE x
[tex]e^y= \frac{x}{e^{x^2} - 1}[/tex]
Find the Natural Log of Both Sides to Make the Left Side "y"
[tex]y = ln(\frac{x}{e^{x^2}-1})[/tex]
Now, FIND THE DERIVATIVE Using Chain Rule!!!
[tex]y' = \frac{1}{\frac{x}{e^{x^2}-1}} * \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)^2}= {\frac{e^{x^2}-1}{x}}* \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)^2} = {\frac{1}{x}}* \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)} = {\frac{1}{x}}* \frac{(e^{x^2}-1 - 2x^2e^{x^2})}{(e^{x^2}-1)}[/tex]
PLEASEEE HELP!!! Let f and g be differentiable functions such that f' (1) = 2 and g’ (1) = 6. If h (x) = 5 f(x) - 4g (x) + 3x^2 - 2, what is the value of h’ (1) ?
The value of h'(1) is -8
Given the following expression;
[tex]h (x) = 5 f(x) - 4g (x) + 3x^2 - 2[/tex]
Differentiating the function above:
[tex]h' (x) = 5 f'(x) - 4g' (x) + 6x[/tex]
Given the following parameters
[tex]f' (1) = 2\\g' (1) = 6[/tex]
Get h'(1)
[tex]h' (1) = 5 f'(1) - 4g' (1) + 6(1)\\h' (1)=5 (2) - 4(6) + 6(1)\\h' (1)= 10-24+6\\h' (1)=-8[/tex]
Hence the value of h'(1) is -8
Learn more here: https://brainly.com/question/1557484
deals with three-dimensional objects, such as cubes and spheres.
Step-by-step explanation:
Length, width and height are three dimensional objects.
I hope this will help you.
what is the answer? I'll give brainiest to whoever is right
Answer:
5) 0 6 ) 1/4
Step-by-step explanation:
So the slope ( gradient ) of line E is 0 because it is straight and it as no slope
The slope of line F is 1/4 because:
Gradient = y^2 - y^1 / x^2 - x^1
Gradient = 2-0 / 6- -2
Gradient= 2 / 8
2/8 in lowest term is 1/4
YOU MIGHT AVE A DIFFERENT METHOD BUT ANSWERS MUST BE THE SAME.
HOPE IT HELPED
what is equivalent to 2(-6c+3)+4c
Answer:
-8c + 6
Step-by-step explanation:
2 (-6c + 3) + 4c
Distribute the 2 with the expression in the parenthesis:
-12c + 6 + 4c
*Parenthesis are no longer needed
Combine Like terms:
-8c + 6
*Ask yourself is it in its simplest form? If no, continue your work. If yes, you have your answer.
A parabola can be drawn given a focus of (-11, -2) and a directrix of x= -3
Write the equation of the parabola in any form.
Check the picture below, so the parabola looks more or less like that.
now, the vertex is half-way between the focus point and the directrix, so that puts it where you see it in the picture, and the horizontal parabola is opening to the left-hand-side, meaning that the distance "P" is negative.
[tex]\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\begin{cases} h=-7\\ k=-2\\ p=-4 \end{cases}\implies 4(-4)[x-(-7)]~~ = ~~[y-(-2)]^2 \\\\\\ -16(x+7)=(y+2)^2\implies x+7=-\cfrac{(y+2)^2}{16}\implies x=-\cfrac{1}{16}(y+2)^2-7[/tex]